A) \[\frac{1+i}{\sqrt{2}}\]
B) \[\cos \frac{\pi }{8}+i\,\sin \frac{\pi }{8}\]
C) \[\frac{1}{4i}\]
D) i
Correct Answer: B
Solution :
\[i{{z}^{4}}=-1\] \[{{z}^{4}}=\frac{-1}{i}\Rightarrow {{z}^{4}}=i\Rightarrow z={{(i)}^{1/4}}\] \[z={{(0+i)}^{1/4}}\] \[z={{\left( \cos \frac{\pi }{2}+i\sin \frac{\pi }{2} \right)}^{1/4}}\] \[z=\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}\] (using De Moivre?s theorem)
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